Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $z = \dfrac{t - 4}{t^2 + 2t - 24} \div \dfrac{-6t + 54}{t + 6} $
Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{t - 4}{t^2 + 2t - 24} \times \dfrac{t + 6}{-6t + 54} $ First factor the quadratic. $z = \dfrac{t - 4}{(t + 6)(t - 4)} \times \dfrac{t + 6}{-6t + 54} $ Then factor out any other terms. $z = \dfrac{t - 4}{(t + 6)(t - 4)} \times \dfrac{t + 6}{-6(t - 9)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (t - 4) \times (t + 6) } { (t + 6)(t - 4) \times -6(t - 9) } $ $z = \dfrac{ (t - 4)(t + 6)}{ -6(t + 6)(t - 4)(t - 9)} $ Notice that $(t - 4)$ and $(t + 6)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ (t - 4)\cancel{(t + 6)}}{ -6\cancel{(t + 6)}(t - 4)(t - 9)} $ We are dividing by $t + 6$ , so $t + 6 \neq 0$ Therefore, $t \neq -6$ $z = \dfrac{ \cancel{(t - 4)}\cancel{(t + 6)}}{ -6\cancel{(t + 6)}\cancel{(t - 4)}(t - 9)} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $z = \dfrac{1}{-6(t - 9)} $ $z = \dfrac{-1}{6(t - 9)} ; \space t \neq -6 ; \space t \neq 4 $